\(\int \frac {(a+b \sin (c+d x^2))^2}{x^2} \, dx\) [21]

Optimal result

Integrand size = 18, antiderivative size = 187 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=-\frac {2 a^2+b^2}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x}+2 a b \sqrt {d} \sqrt {2 \pi } \cos (c) \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+b^2 \sqrt {d} \sqrt {\pi } \cos (2 c) \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-2 a b \sqrt {d} \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)+b^2 \sqrt {d} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)-\frac {2 a b \sin \left (c+d x^2\right )}{x} \] Output:

-1/2*(2*a^2+b^2)/x+1/2*b^2*cos(2*d*x^2+2*c)/x+2*a*b*d^(1/2)*2^(1/2)*Pi^(1/ 
2)*cos(c)*FresnelC(d^(1/2)*2^(1/2)/Pi^(1/2)*x)+b^2*d^(1/2)*Pi^(1/2)*cos(2* 
c)*FresnelS(2*d^(1/2)*x/Pi^(1/2))-2*a*b*d^(1/2)*2^(1/2)*Pi^(1/2)*FresnelS( 
d^(1/2)*2^(1/2)/Pi^(1/2)*x)*sin(c)+b^2*d^(1/2)*Pi^(1/2)*FresnelC(2*d^(1/2) 
*x/Pi^(1/2))*sin(2*c)-2*a*b*sin(d*x^2+c)/x
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\frac {-2 a^2-b^2+b^2 \cos \left (2 \left (c+d x^2\right )\right )+4 a b \sqrt {d} \sqrt {2 \pi } x \cos (c) \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+2 b^2 \sqrt {d} \sqrt {\pi } x \cos (2 c) \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-4 a b \sqrt {d} \sqrt {2 \pi } x \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)+2 b^2 \sqrt {d} \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)-4 a b \sin \left (c+d x^2\right )}{2 x} \] Input:

Integrate[(a + b*Sin[c + d*x^2])^2/x^2,x]
 

Output:

(-2*a^2 - b^2 + b^2*Cos[2*(c + d*x^2)] + 4*a*b*Sqrt[d]*Sqrt[2*Pi]*x*Cos[c] 
*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x] + 2*b^2*Sqrt[d]*Sqrt[Pi]*x*Cos[2*c]*Fresne 
lS[(2*Sqrt[d]*x)/Sqrt[Pi]] - 4*a*b*Sqrt[d]*Sqrt[2*Pi]*x*FresnelS[Sqrt[d]*S 
qrt[2/Pi]*x]*Sin[c] + 2*b^2*Sqrt[d]*Sqrt[Pi]*x*FresnelC[(2*Sqrt[d]*x)/Sqrt 
[Pi]]*Sin[2*c] - 4*a*b*Sin[c + d*x^2])/(2*x)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3884, 6, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx\)

\(\Big \downarrow \) 3884

\(\displaystyle \int \left (\frac {a^2}{x^2}+\frac {2 a b \sin \left (c+d x^2\right )}{x^2}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^2}+\frac {b^2}{2 x^2}\right )dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x^2}+\frac {2 a b \sin \left (c+d x^2\right )}{x^2}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2+b^2}{2 x}+2 \sqrt {2 \pi } a b \sqrt {d} \cos (c) \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-2 \sqrt {2 \pi } a b \sqrt {d} \sin (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {2 a b \sin \left (c+d x^2\right )}{x}+\sqrt {\pi } b^2 \sqrt {d} \sin (2 c) \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+\sqrt {\pi } b^2 \sqrt {d} \cos (2 c) \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x}\)

Input:

Int[(a + b*Sin[c + d*x^2])^2/x^2,x]
 

Output:

-1/2*(2*a^2 + b^2)/x + (b^2*Cos[2*c + 2*d*x^2])/(2*x) + 2*a*b*Sqrt[d]*Sqrt 
[2*Pi]*Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x] + b^2*Sqrt[d]*Sqrt[Pi]*Cos[2* 
c]*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]] - 2*a*b*Sqrt[d]*Sqrt[2*Pi]*FresnelS[Sq 
rt[d]*Sqrt[2/Pi]*x]*Sin[c] + b^2*Sqrt[d]*Sqrt[Pi]*FresnelC[(2*Sqrt[d]*x)/S 
qrt[Pi]]*Sin[2*c] - (2*a*b*Sin[c + d*x^2])/x
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3884

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.72

method	result	size
parts	\(-\frac {a^{2}}{x}+b^{2} \left (-\frac {1}{2 x}+\frac {\cos \left (2 d \,x^{2}+2 c \right )}{2 x}+\sqrt {d}\, \sqrt {\pi }\, \left (\cos \left (2 c \right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )+\sin \left (2 c \right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )\right )\right )+2 a b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{x}+\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )-\sin \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )\right )\)	\(134\)
default	\(-\frac {a^{2}+\frac {b^{2}}{2}}{x}-\frac {b^{2} \left (-\frac {\cos \left (2 d \,x^{2}+2 c \right )}{x}-2 \sqrt {d}\, \sqrt {\pi }\, \left (\cos \left (2 c \right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )+\sin \left (2 c \right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )\right )\right )}{2}+2 a b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{x}+\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )-\sin \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )\right )\)	\(137\)
risch	\(\frac {i b^{2} d \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i d}\, x \right ) {\mathrm e}^{-2 i c}}{4 \sqrt {i d}}-\frac {i b^{2} d \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i d}\, x \right ) {\mathrm e}^{2 i c}}{2 \sqrt {-2 i d}}+\frac {a b d \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i d}\, x \right ) {\mathrm e}^{i c}}{\sqrt {-i d}}+\frac {a b d \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {i d}\, x \right ) {\mathrm e}^{-i c}}{\sqrt {i d}}-\frac {a^{2}}{x}-\frac {b^{2}}{2 x}-\frac {2 a b \sin \left (d \,x^{2}+c \right )}{x}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{2 x}\)	\(172\)

Input:

int((a+b*sin(d*x^2+c))^2/x^2,x,method=_RETURNVERBOSE)
 

Output:

-1/x*a^2+b^2*(-1/2/x+1/2/x*cos(2*d*x^2+2*c)+d^(1/2)*Pi^(1/2)*(cos(2*c)*Fre 
snelS(2*d^(1/2)*x/Pi^(1/2))+sin(2*c)*FresnelC(2*d^(1/2)*x/Pi^(1/2))))+2*a* 
b*(-sin(d*x^2+c)/x+d^(1/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelC(d^(1/2)*2^(1/ 
2)/Pi^(1/2)*x)-sin(c)*FresnelS(d^(1/2)*2^(1/2)/Pi^(1/2)*x)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\frac {2 \, \sqrt {2} \pi a b x \sqrt {\frac {d}{\pi }} \cos \left (c\right ) \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) - 2 \, \sqrt {2} \pi a b x \sqrt {\frac {d}{\pi }} \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \left (c\right ) + \pi b^{2} x \sqrt {\frac {d}{\pi }} \cos \left (2 \, c\right ) \operatorname {S}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) + \pi b^{2} x \sqrt {\frac {d}{\pi }} \operatorname {C}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) \sin \left (2 \, c\right ) + b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2}}{x} \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^2,x, algorithm="fricas")
 

Output:

(2*sqrt(2)*pi*a*b*x*sqrt(d/pi)*cos(c)*fresnel_cos(sqrt(2)*x*sqrt(d/pi)) - 
2*sqrt(2)*pi*a*b*x*sqrt(d/pi)*fresnel_sin(sqrt(2)*x*sqrt(d/pi))*sin(c) + p 
i*b^2*x*sqrt(d/pi)*cos(2*c)*fresnel_sin(2*x*sqrt(d/pi)) + pi*b^2*x*sqrt(d/ 
pi)*fresnel_cos(2*x*sqrt(d/pi))*sin(2*c) + b^2*cos(d*x^2 + c)^2 - 2*a*b*si 
n(d*x^2 + c) - a^2 - b^2)/x
 

Sympy [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x^{2}}\, dx \] Input:

integrate((a+b*sin(d*x**2+c))**2/x**2,x)
 

Output:

Integral((a + b*sin(c + d*x**2))**2/x**2, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=-\frac {\sqrt {d x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, d x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, d x^{2}\right )\right )} \cos \left (c\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, d x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b}{4 \, x} - \frac {{\left (\sqrt {2} \sqrt {d x^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 2 i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 2 i \, d x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} + 8\right )} b^{2}}{16 \, x} - \frac {a^{2}}{x} \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^2,x, algorithm="maxima")
 

Output:

-1/4*sqrt(d*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, I*d*x^2) - (I + 1)*sqrt(2)* 
gamma(-1/2, -I*d*x^2))*cos(c) + ((I + 1)*sqrt(2)*gamma(-1/2, I*d*x^2) - (I 
 - 1)*sqrt(2)*gamma(-1/2, -I*d*x^2))*sin(c))*a*b/x - 1/16*(sqrt(2)*sqrt(d* 
x^2)*((-(I + 1)*sqrt(2)*gamma(-1/2, 2*I*d*x^2) + (I - 1)*sqrt(2)*gamma(-1/ 
2, -2*I*d*x^2))*cos(2*c) + ((I - 1)*sqrt(2)*gamma(-1/2, 2*I*d*x^2) - (I + 
1)*sqrt(2)*gamma(-1/2, -2*I*d*x^2))*sin(2*c)) + 8)*b^2/x - a^2/x
 

Giac [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\int { \frac {{\left (b \sin \left (d x^{2} + c\right ) + a\right )}^{2}}{x^{2}} \,d x } \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^2,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x^2 + c) + a)^2/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x^2} \,d x \] Input:

int((a + b*sin(c + d*x^2))^2/x^2,x)
 

Output:

int((a + b*sin(c + d*x^2))^2/x^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^2} \, dx=\frac {\left (\int \frac {\sin \left (d \,x^{2}+c \right )^{2}}{x^{2}}d x \right ) b^{2} x +2 \left (\int \frac {\sin \left (d \,x^{2}+c \right )}{x^{2}}d x \right ) a b x -a^{2}}{x} \] Input:

int((a+b*sin(d*x^2+c))^2/x^2,x)
 

Output:

(int(sin(c + d*x**2)**2/x**2,x)*b**2*x + 2*int(sin(c + d*x**2)/x**2,x)*a*b 
*x - a**2)/x